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Proof step 8 from proof sum-swap (previous step)

∀ n, nnf. Σ(n, λ k0. Σ(m0, λ k1. nnf(k0, k1))) = Σ(m0, λ k1. Σ(n, λ k0. nnf(k0, k1))) ⊦ Σ(n, λ n. nnf(n, m0)) + Σ(n, λ n. Σ(m0, λ k1. nnf(n, k1))) = Σ(n, λ k0. nnf(k0, m0)) + Σ(m0, λ k1. Σ(n, λ k0. nnf(k0, k1)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ n, nnf. Σ(n, λ k0. Σ(m0, λ k1. nnf(k0, k1))) = Σ(m0, λ k1. Σ(n, λ k0. nnf(k0, k1))) ⊦ Σ(n, λ n. Σ(m0, λ k1. nnf(n, k1))) = Σ(m0, λ k1. Σ(n, λ k0. nnf(k0, k1)))

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2